The Definite Integral
Key Questions
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Answer:
I don't have a one sentence answer
Explanation:
Above the x-axis
If#f(x)# is non-negative over the interval#[a,b]# , then
#int_a^b f(x) dx# is the area of the region between the graph and the#x# -axis between#x=a# and#x=b# Below the x-axis
If#f(x)# is non-positive over the interval#[a,b]# , then
#int_a^b f(x) dx# is the negative of area of the region between the graph and the#x# -axis between#x=a# and#x=b# Example:
#int_0^3 sqrt(9-x^2) dx# .#a=0# and#b=3# .
#f(x) = sqrt(9-x^2)# is never negative, so it is not negative on#[0,3]# Therefore the integral is equal to the area under the curve and above the
#x# -axis between#x=0# and#x=3# .The graph of
#y = sqrt(9-x^2)# is the part of#y^2 = 9-x^2# that has non-negative#y# -values. It is the upper semicircle for#x^2+y^2 = 9#
The part between#x=0# and#x=3# is a quarter of a circle with radius 3.graph{y = sqrt(9-x^2)*(sqrt(1.5^2-(x-1.5)^2))/(sqrt(1.5^2-(x-1.5)^2)) [-2.63, 6.137, -0.812, 3.572]}
So
#int_0^3 sqrt(9-x^2) dx# is#1/4# of the area of the circle with radius#3#
#int_0^3 sqrt(9-x^2) dx = (9 pi)/4# 
Jim H
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Jun 19 2015
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Answer:
An indefinite integral of a function
#f(x)# is a family of functions#g(x)# for which:#g'(x)=f(x)# Explanation:
An indefinite integral of a function
#f(x)# is a family of functions#g(x)# for which:#g'(x)=f(x)# .Examples:
1) if
#f(x)=x^3# , then indefinite integral is:#int x^3dx=x^4/4+C# , because:#(x^4/4)'=4*x^3/4=x^3# , and#C'=0# for any real constant#C# 2) If
#f(x)=cosx# , then#int cosx dx= sinx+C# , because:#(sinx)'=cosx# Daniel L. · 1 · Aug 31 2015
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A definite integral is when you integrate a function over a specified interval. When completed you have a definite answer.
Definite Integral because it is bounded
#int_0^1 3xdx# evaluates to#[(3(1)^2)/2 - (3(0)^2)/2] = 3/2 - 0 = 3/2=1.5# Indefinite Integral because it is not bounded
#int 3x dx# evaluates to#(3x^2)/2 + C# 
AJ Speller
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Sep 10 2014